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Q.1 Two angles of a triangle measure $58^{\circ} \ and\ 87^{\circ}$. Find the measure of the third angle.

Sum of all the angles of a triangle $= 180^{\circ}$

Let the third angle $= x$

Therefore $58 + 87 + x = 180$

Or $x= 35^{\circ}$

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Q.2. Three angles of a triangle are in the ratio of 2 : 5 : 8.  Find the angles of the triangle.

Let the angles be $2x, 5x, \ and\ 8x$

Therefore $2x+5x+8x = 180$

Or $x= 12$

Therefore the angles are $24^{\circ}, 60^{\circ}, \ and\ 96^{\circ}.$

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Q.3. One angle of the triangle is $76$ and the other two angles are equal. Find the measure of each angle of the triangle.

Let the two equal angles be $x^{\circ}$

Therefore $76 + 2x = 180$

Or $x = 52$

Therefore the angles are $76^{\circ}, 52^{\circ} \ and\ 52^{\circ}.$

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Q.4. One angle of the triangle is $85^{\circ}$ and the other two angles are in the ratio $8 :11$Find the measures of each of the unknown angles.

Let the two unknown angles be $8x \ and\ 11 x$

Therefore $85+8x+11x = 180$

Or $19x = 95$

Or $x = 5$

Therefore the two angles are $40^{\circ} \ and\ 55^{\circ}.$

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Q.5. Three angles of a triangle measures $(2x)^{\circ}, (4x-7)^{\circ} \ and\ (5x-11)^{\circ}$.  Find the value of $x$ and the measures of all the angles.

$2x + (4x-7) + (5x-11) = 180$

Or $x = 18$

Hence the angles are $36^{\circ}, 65^{\circ}, \ and\ 79^{\circ}$

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Q.6. In $\Delta ABC, \angle A=x^{\circ}, \angle B=(2x-15)^{\circ}, \ and\ \angle C=(3x+21)^{\circ}$. Find the value of $x$ and the measure of all the angles of the triangle.

$x+(2x-15)+(3x+21)=180$

or $x=29$

Hence $\angle A=29^{\circ}, \angle B=43^{\circ}, \ and\ \angle C=108^{\circ}$

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Q.7. In $\Delta ABC, \angle A=(2x+15)^{\circ}, \angle B=x^{\circ}, \ and\ \angle C=(3x-15)^{\circ}$. Find the value of x and the measure of all the angles of the triangle and show that the triangle is an isosceles triangle.

$(2x+15)+x+(3x-15)=180 \ or\ x=30$

Hence $\angle A=75^{\circ}, \angle B=30^{\circ}, \ and\ \angle C=75^{\circ}.$

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Q.9. In $\Delta ABC, \angle A=3\ \angle B=6 \ \angle C$ . Find the angles of the triangle.

Let $\angle A=3\angle B=6 \angle C=6x$

$\Rightarrow \angle A=6x, \angle B=2x, and \angle C=x$

$\Rightarrow 6x+2x+x=180 \ or\ x=20$

$\Rightarrow \angle A=120^{\circ}, \angle B=40^{\circ} \ and\ \angle C=20^{\circ}$

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Q.10. In the adjoining figure, it is given that $\angle ABC=40^{\circ}, \angle ACD=70^{\circ}, \angle ACB=x^{\circ} \ and\ \angle EAC=y$. Find the value of $latex $x and y$. Answer: $x+70=180 \ hence\ x=110^{\circ}$ Note: Sum of angles of adjacent angles on a straight line $=180$ $\Rightarrow 40+110+(180-y)=180$ Note:Sum of all angles of a triangle $=180$ $\Rightarrow y=150^{\circ}$ $\\$ Q.11. In the adjoining figure, it is given that $\angle A=50^{\circ} \ and\ \angle B=70^{\circ}$ and $BO \ and\ CO$ are bisectors of $\angle B \ and\ \angle C$ respectively. Find the measure of $\angle BOC.$ Answer: Let $\angle OCA= \angle OCB=x^{\circ}$ Since $BO \ bisects \angle A=ABC, \angle ABO= \angle OBC=35^{\circ}.$ $\Rightarrow 50+70+2x=180$ $\Rightarrow x=30^{\circ}=\angle OCB$ $\Rightarrow 35+30+ \angle BOC=180$ $\Rightarrow \angle BOC=115^{\circ}$ $\\$ Q.12. Using the information given in the adjoining figure, find the value of $x \ and\ y.$ Answer: $36+y+x=180$ $(24+36)+2x+64=180$ $\Rightarrow x=28^{\circ}$ $\Rightarrow 36+y+28=180 \ or\ y=116^{\circ}$ $\\$ Q.13. In the adjoining figure, DCBE is a straight line, $\angle ABD=135^{\circ} \ and\ \angle ACE=125^{\circ}$. Find $\angle BAC.$ Answer: $\angle ABC=180-135=45^{\circ}$ $\angle ACB=180-125=55^{\circ}$ Therefore $\angle BAC=180- \angle ABC- \angle ACB$ $=180-45-55=80$ $\\$ Q.14. Find, giving reasons, the unknown angles marked by $x, \ y$ in the following figures. Answer: i) $58+x+(180-123)=180$ $\Rightarrow x=65^{\circ}$ $\\$ ii) $42+y=180$ $\Rightarrow y=138^{\circ}$ $\\$ iii) $x+72+42=180$ $\Rightarrow x=68^{\circ}$ $\\$ iv) $50+x+x=180$ $\Rightarrow x=65^{\circ}$ $\\$ v) $65+y=180$ $\Rightarrow y=115^{\circ}$ $\\$ vi) $30+y+110=180$ $\Rightarrow y=40^{\circ}$ Given $AC=AD$ which means $\angle ACD= \angle ADC=70^{\circ}$ $x+70+70=180$ $\Rightarrow x=40^{\circ}$ $\\$ Q.15. In the adjoining figure, if $\angle A=(x+5)^{\circ}, \angle B=(2x+3)^{\circ} \ and\ \angle BCD=(5x-14)^{\circ}$, find the value of $x$ and hence, find the measure of $i)\ \angle BCD \ ii)\ \angle ACB$ Answer: $\angle A+ \angle B+ \angle C=180^{\circ}$ $\Rightarrow (x+5)+(2x+3)+{180-(5x-14)}=180$ $\Rightarrow x=11 \ Therefore\ \angle A=16^{\circ}, \angle B=25^{\circ}$ $\Rightarrow i) \angle BCD=41^{\circ} and ii) \angle ACB=180-41=139^{\circ}$ $\\$ Q.16. In $\Delta ABC$, side AC has been produced to D. If $\angle BCD=125^{\circ} \ and\ \angle A : \angle B=2 :3$, find the measure of $\angle A \ and\ \angle B.$ Answer: Let $\angle A=2x \ and\ \angle B=3x.$ $\Rightarrow 2x+3x+(180-125)=180$ $\Rightarrow x=25$ Therefore $\angle A=50^{\circ} \ and\ \angle B=75^{\circ}$ $\\$ Q.17. In the adjoining figure, it is given that: $BC \parallel DE,\ \angle BAC=35 \ and\ \angle BCE=102$. Find the measure of $i) \angle ABC ii)\ \angle ADE \ and\ iii) \angle CED$ Answer: i) $\angle ABC+35+(180-102)=180$ $\Rightarrow \angle ABC=67^{\circ}$ $\angle DEA+102=180$ Note:Co-interior Angles $\Rightarrow \angle DEA=78^{\circ}$ ii) $\angle ADE+78+35=180$ $\Rightarrow \angle ADE=67^{\circ}$ iii) $\angle CED+67+35=180$ $\Rightarrow \angle CED=78^{\circ}$ $\\$ Q.18. In the adjoining figure, it is given that:$AB=AC, \angle BAC=36^{\circ}, \angle ADB=45^{\circ} and \angle AEC=40^{\circ}$. Find $i) \angle ABC, ii) \angle ACB iii) \angle DAB and iv) \angle EAC$ Answer: $AB=AC \Rightarrow \angle ABC= \angle ACB=x$ Therefore $36+2x=180\Rightarrow x=72$ i) & ii) Hence $\angle ABC= \angle ACB=72^{\circ}$ iv) $\angle CAE+(180-72)+40=180$ $\Rightarrow \angle CAE=32^{\circ}$ iii) $\angle DAB+(180-72)+45=180$ $\Rightarrow \angle DAB=27^{\circ}$ $\angle DAE=27+36+32=95^{\circ}$ $\\$ Q.19. Using the information given in the adjoining figure, calculate the values of $x \ and\ y.$ Answer: In $\Delta ABC$ $34+(180-x)+62=180$ $\Rightarrow x=96^{\circ}$ In $\Delta DCE 24+96+(180-y)=180$ $\Rightarrow y=120^{\circ}$ $\\$ Q.20. Sides $AB \ and\ AC of \Delta ABC$ have been produced to D and E respectively. $BI and CI$ are bisectors of $\angle CBD and \angle BCE$ respectively. If $\angle BAC=70^{\circ} and \angle ACB 50^{\circ}$, find the measure of $\angle BIC$ Answer: Given $\angle DBI= \angle IBC=x and \angle BCI= \angle ICE=y$ $\therefore 50+2y=180\Rightarrow y=65$ $\angle ABC+50+70=180$ $\Rightarrow \angle ABC=60^{\circ}$ $60+2x=180$ $\Rightarrow x=60$ $\therefore 60+ \angle BIC+65=180$ $\Rightarrow \angle BIC=55^{\circ}$ $\\$ Q.21. In the adjoining figure, it is being given that $BC \parallel DE, \angle CED=70^{\circ}, \angle CBA=84^{\circ} \ and\ \angle BAC=x^{\circ}$. Find the value of $x$. Answer: $\angle DEC= \angle BCA=70^{\circ}$ Note:Alternate angles $x+84+70=180 \Rightarrow x=26^{\circ}$ $\\$ Q.22. In the adjoining figure, it is being given that $BC \parallel DE, \angle EDC=25^{\circ}, \angle ECD=20^{\circ}, \angle ABC=70^{\circ} \angle BAC=x^{\circ} and \angle DEA=y^{\circ}$. Find the value of $x \ and\ y$. Answer: In $\Delta DEC,$ $25+20+ \angle DEC=180$ $Rightarrow \angle DEC=135^{\circ}$ $\therefore y+135=180$ $\Rightarrow y=45^{\circ}$ $BC \parallel DE \Rightarrow \angle CBD= \angle EDA=70^{\circ}$ In $\Delta ADE,$ $70+45+x=180$ $\Rightarrow x=65^{\circ}$ $\\$ Q.23. In the adjoining figure, it is given that $\angle ACB=42^{\circ}, \angle DEB= \angle CAB=90^{\circ}, \angle ABC=x^{\circ}, \angle BDE=y^{\circ} \ and\ \angle AFE=z^{\circ}.$ Find the value of $x, \ y \ and\ z.$ Answer: In$latex \Delta FEC, 90+42+(180-z)=180 &s=1\$

$\Rightarrow z=132^{\circ}$

In $\Delta ADF, (180-132)+90+y=180$

$\Rightarrow y=42^{\circ}$

In $\Delta DBE, 42+90+x=180$

$\Rightarrow x=48^{\circ}$

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Q.24. In the adjoining figure, $AE \parallel BC, \angle DAE=x^{\circ}, \angle ACB=(x-15)^{\circ}, \angle BAC=(\frac{x}{2}+y)^{\circ} \ and\ \angle ABC=(y+15)^{\circ}$

In $\Delta ABC, (y+15)+(x-15)+(\frac{x}{2}+y)=180$

$\Rightarrow 2y+\frac{3x}{2}=180 ... ... ... ...(i)$

Since $AE \parallel BC,$

$x=(y+15) ... ... ... ...ii)$ Note:Alternate Angles

Substituting ii) in i) we get

$2y+\frac{3}{2} (y+15)=180$

$\Rightarrow 4y+3y+45=360\Rightarrow y=45^{\circ}$

Substituting the value of $y \ in \ i) x=45+15=60^{\circ}$

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Q.25. In the adjoining figure, $BA \parallel CD$. Find the values of $x \ and\ y.$

In $\Delta ABC$

$(y+7)+(y-10)+(4x-5)=180$

$\Rightarrow 2y+4x=188 ... ... ... ... (i)$

Since $BA \parallel CD,$

$(y-10)=5x ... ... ... ...(ii)$

Substituting ii) in i) we get

$2(5x+10)+4x=188$

$\Rightarrow 14x=168$

$\Rightarrow x= 12^{\circ}$

Substituting in ii) we get     $y=70^{\circ}$

E Board: Suggested Books     ICSE Board:  Foundation Mathematics
Class 8: Reference Books     Class 8: NTSE Preparation
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